Module compile_type_eq::how_it_works

Learn how comparison works.

Core of library is following lines:

struct NotEq;
struct Eq;

trait CompileEq<T, Result> {}
impl<T> CompileEq<T, Eq> for T {}
impl<T, U> CompileEq<U, NotEq> for T {}

Than we creating a function for checking:

fn assert_types_not_eq<T, U, Proof>()
where
    T: CompileEq<U, Proof>,
{ }

Why it works? Suppose we are passing to a function two different types T and U. Let's go find implementations of CompileEq for T, as if we were a compiler. We can find only one implementation: impl<T, U> CompileEq<U, NotEq> for T {}. Because there are no other candidats, compiler substite this implementation.

Than suppose that we have passed the same type T to a function twice. Let's go find implementations of CompileEq for T, as if we were a compiler. We can find two implementations: impl<T> CompileEq<T, Eq> for T {} and impl<T, U> CompileEq<U, NotEq> for T {}. Because there are two different impls, the compiler generates a type inference error.